SECONDARY STORAGE DEVICES: MAGNETIC TAPES AND CD-ROM

SECONDARY STORAGE DEVICES: MAGNETIC TAPES AND CD-ROM

Contents of today s lecture: Magnetic Tapes Characteristics of magnetic tapes Data organization on 9-track tapes Estimating tape length requirements Estimating data transmission times Disk versus tape CD-ROM Physical Organization of CD-ROM CD-ROM Strengths and Weaknesses Reference: Folk, Zoellick and Riccardi, File Structures, 1998. Section 3.2, 3.5 and 3.6. Lucia Moura 53

Characteristics of Magnetic Tapes No direct access, but very fast sequential access. Resistant to different environmental conditions. Easy to transport, store, cheaper than disk. Before, it was widely used to store application data; nowadays, it s mostly used for backups or archives (tertiary storage). Lucia Moura 54

Data Organization on Nine-Track Tapes In a tape, the logical position of a byte within a file is the same as its physical position in the file (sequential access). Nine-track tape: 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 ----------------------------------------------------- <-Gap-> <-- Data Block --> <-Gap-> - Data blocks are separated by interblock GAPS. - 9 parallel tracks (each is a sequence of bits) - A frame is a 1-bit slice of the tape corresponding to 9 bits (one in each track) which correspond to 1 byte plus a parity bit. In the example above, the byte stored in the frame that is shown is: 01101001. The parity bit is 1, since we are using odd parity, i.e., the total number of bits is odd. Lucia Moura 55

Complete the parity bit in the examples below: 11111111 00000000 00100000 Since 000000000 cannot correspond to a valid byte, this is used to mark the interblock gap. So, if we say that this tape has 6,250 bits per inch (bpi) per track, indeed it stores 6,250 bytes per inch when we take into account the 9 tracks. Lucia Moura 56

Estimating Tape Length Requirements Performance of tape drives can be measured in terms of 3 quantities: - Tape density = 6250 bpi (bits per inch per track) - Tape speed = 200 inches per second (ips) - Size of interblock gap = 0.3 inch File characteristics: - Number of records = 1,000,000 - Size of record = 100 bytes How much tape is needed? It depends on the blocking factor (how many records per data block). Let us compute the space requirement in two cases: A) Blocking factor = 1 B) Blocking factor = 50 Space requirement ( s ) b = length of data block (in inches) g = length of interblock gap (in inches) n = number of data blocks s = n x (b + g) Lucia Moura 57

A) Blocking factor = 1 b = block size/tape density = 100 bytes/6250 bpi= 0.016 inch n = 1,000,000 (recall blocking factor is 1) s = 1,000,000 x (0.016 + 0.3) inch = 316,000 inches 26,333 feet (Absurd to have the length of the data block smaller than the interblock gap!) B) Blocking factor = 50 b = 50 x 100 bytes/6,250 bpi = 0.8 inch n = 1,000,000 records/50 records per block = 20,000 blocks s = 20,000 x (0.8 + 0.3) inch = 22,000 inches 1,833 feet An enormous saving by just choosing a higher blocking factor. Effective Recording Density (ERD) ERD = number of bytes per block / number of inches to store a block In previous example : A) Blocking factor =1: E.R.D. = 100/0.316 316.4 bpi B) Blocking factor =50: E.R.D. = 5,000/1.1 4,545.4 bpi The Nominal Density was 6,250 bpi! Lucia Moura 58

Estimating Data Transmission Times Nominal Rate = tape density (bpi) x tape speed (ips) In a 6,250 - bpi, 200 - ips tape : Nominal Rate = 6,250 bytes/inch x 200 inches/second = = 1,250,000 bytes/sec 1,250 KB/sec Effective Transmission Rate = E.R.D. x tape speed In the previous example: A) E.T.R. = 316.4 bytes/inch x 200 inches/sec = 63,280 bytes/sec 63.3 KB/sec B) E.T.R. = 4,545.4 bytes/inch x 200 inches/sec = 909,080 bytes/sec 909 KB/sec Note : There is a tradeoff between increasing blocking factor for increasing speed & space utilization and decreasing it for reducing the size of the I/O buffer. Disk versus Tape In the past : Disks and Tapes were used for secondary storage: disks preferred for random access and tapes for sequential access. Now : Disks have taken over most of secondary storage (lower cost of disk and lower cost of RAM which allows large I/O buffer). Tapes are mostly used for tertiary storage. Lucia Moura 59

Physical Organization of CD-ROM Compact Disc - read only memory (write once) Data is encoded and read optically with a laser Can store around 600 MB data Digital data is represented as a series of Pits and Lands. Pit = a little depression, forming a lower level in the track Land = the flat part between pits, or the upper levels in the track Reading a CD is done by shining a laser at the disc and detecting changing reflections patterns. 1 = change in height (land to pit or pit to land) 0 = a fixed amount of time between 1 s LAND PIT LAND PIT LAND... ------------+ +-------------+ +----...... 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0... Changes in height in the track are detected as changes of intensity of the reflected light. Note: We cannot have two 1 s in a row! Lucia Moura 60

Indeed, because of other limitations there must be at least two and at most ten 0 s between two 1 s. Therefore, each of the 256 bytes must be encoded into a sequence of bits that has every pair of 1 s separated by at least two zeros. There are exactly 267 binary words of length 14 that satisfy this property; 256 of them were chosen to represent every possible byte in the so-called eight to fourteen modulation. We could not encode bytes using 13 bits since there are only 188 words of length 13 having the desired property. Eight to fourteen modulation (EFM) encoding table: Decimal Original Translated Value Bits Bits 0 00000000 01001000100000 1 00000001 10000100000000 2 00000010 10010000100000 3 00000011 10001000100000 4 00000100 01000100000000 5 00000101 00000100010000 6 00000110 00010000100000 7 00000111 00100100000000 8 00001000 01001001000000......... Note that: Since 0 s are represented by the length of time between transitions, we must travel at constant linear velocity on the tracks. Lucia Moura 61

Comparing CD-ROM with magnetic disks CR-ROM CLV = Constant Linear Velocity Sectors organized along a spiral Sectors have same linear length (data packed at its maximum density permitted) Advantage: takes advantage of all storage space available Disadvantage: has to change rotational speed when seeking (slower towards the outside) Magnetic Disks CAV = Constant Angular Velocity Sectors organized in concentric track Sectors have same angular length (data written less densely in the outer tracks) Advantage: operates on constant speed, timing marks to delimit tracks Disadvantage: it doesn t use up all storage available Lucia Moura 62

CD-ROM addressing and poor Seek performance Addressing 1 second of play time is divided up into 75 sectors. Each sector holds 2KB. 60 Min CD : 60 min x 60 sec/min x 75 sectors/sec = 270,000 sectors = 540,000 KB 540 MB A sector is addressed by : Minute : Second : Sector 16:22:34 16 min, 22 sec, 34th sector Difficulty in Seeking To read address of a sector it must be at the correct speed But knowing the correct speed depends on the ability to read the address info! The drive control mechanism solves this problem by trial-and-error. This slows down the performance! Lucia Moura 63

CD-ROM Strength and Weaknesses Seek performance ( 500 msecs ) - Slow Our old analogy : 20 secs (RAM) 58 days (Magnetic Disks) 2.5 years (CD-ROM) Data transfer rate - 150 KB/sec - Slow (while 3,000 KB/sec for magnetic disks), but 5 times faster than floppy disks. Storage capacity is 600 MB; good for storing texts. Read-only access (publishing medium). File structure designer can take advantage of that. Things changed nowadays : Most drives use CAV or combination of CAV and CLV Other types of compact discs : - CD-R = compact disc-recordable - CD-RW = compact disc-rewritable They use different technologies which simulates the effect of Pits and Lands. Lucia Moura 64